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quiz8sol - November14,2007 PHY2053 Quiz8(Chapter8 Name:UFID...

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November 14, 2007 PHY2053  Quiz 8 (Chapter 8) Name:                                   UFID: 1. (5 pts) A 20.0m, 40.0kg uniform ladder rests against a frictionless wall, making an  angle of 60º with the horizontal. Find the minimum coefficient of static friction between  ladder and ground so that a 80.0 kg firefighter can reach the top of the ladder. Since each component of the net force equals zero, we get X: n-mg-Mg = 0   n = (m+M)g = (80+40)9.8 = 1176 N, Y: N-f = 0   f = N,  where m = 80kg, M = 40kg, n/N are the normal force from the ground/the wall.  Now we calculate the torque around the bottom of the ladder when the firefighter   reaches the top: mgLcos +Mg(L/2)cos -NLsin  = 0  θ θ θ  N = (gcos (m+M/2))/sin  = 9.8cos60º(80+40/2)/sin60º = 565.8 N θ θ
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